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Proof step 24 from proof square-isSquare (previous step)

isSquare(mkDecimal(j * j, 2 * abs(abs(i0) - 1)).int), ∀ i1, i0. i→n(i1) < i→n(j * 10) → true → isSquare(mkDecimal(i1 * i1, 2 * abs(i0)).int) ⊦ i0 = 0, j = 0, isSquare(mkDecimal(j * j, 2 * abs(i0) - 2).int)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
isSquare(mkDecimal(j * j, 2 * (abs(i0) - 1)).int), ∀ i1, i0. i→n(i1) < i→n(j * 10) → true → isSquare(mkDecimal(i1 * i1, 2 * abs(i0)).int) ⊦ i0 = 0, j = 0, isSquare(mkDecimal(j * j, 2 * abs(i0) - 2).int)

Rule argument

Additional information

Used simplifier rules: abs-abs: ⊦ j ≠ 0 → abs(abs(j) - 1) = abs(j) - 1; from specification int-abs